Gọi oxit kim loại là $XO$
$XO + 2HCl \to XCl_2 + H_2O\\n_{XO}=\dfrac{7,2}{M_X+16}\,(mol)\\n_{XCl_2}=\dfrac{12,7}{M_X+71}\\\text{Ta có: }n_{XO}=n_{XCl_2}\\\to \dfrac{7,2}{M_X+16}=\dfrac{12,7}{M_X+71}\\\to 7,2(M_X+71)=12,7(M_X+16)\\\to 7,2M_X+511,2=12,7M_X+203,2\\\to 5,5M_X=308\\\to M_X=56\\\to X\text{ là }Fe \\\to\text{Công thức oxit là }FeO$